Finding the points at which a surface has horizontal tangent planes
Find the points at which the surface $$ x^2 +2y^2+z^2 -2x -2z -2 = 0 $$ has horizontal tangent planes. Find the equation of these tangent planes.
I found that $$ \nabla f = (2x-2,4y) $$ I'm thinking that the gradient vector must be equal to $(0,0)$ so $x = 1, y=0$ which implies $z = 3,-1$. So the points are $(1,0,3)$ and $(1,0,-1)$. How would we write the equation of these tangent planes?
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$\begingroup$Your answer is correct, but the logic is wrong. You should consider the above surface as a level surface of the function:
$$ f(x, y, z) = x^2 + 2y^2 + z^2 - 2x - 2z - 2$$
Then the gradient should be a vector with 3 components.
$$ \nabla f(x, y, z) = (2x - 2, 4y, 2z - 2) $$
If the tangent plane is horizontal, the gradient must point in the $z$-direction, therefore the $x$ and $y$ components are $0$. So it follows that $x = 1$ and $y = 0$
Finally, the equations are just $z = 3$ and $z = -1$ since horizontal planes have the same $z$ coordinates everywhere.
Another way is to treat $z$ as a function of $x$ and $y$, then set $\partial z / \partial x$ and $\partial z / \partial y$ equal to $0$. However, those expressions are not the same as what you wrote, since they have to be found through implicit differentiation. I'm not sure which method you were trying to use.
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